Get Ahead in Geometric Probability: Expert Guidance and Real-World Applications

At a given distance from the origin, which convex subsets of an ${\mathrm{\ell <!--\; \ell \; -->}}_p$-ball have the maximal volume?
For a positive integer n and $p\in <!--\; \in \; -->[1,\mathrm{\infty <!--\; \infty \; -->}]$, let ${\mathbb{B}}_{n,p}:=\{x\in <!--\; \in \; -->{\mathbb{R}}^n\mid <!--\; \mid \; -->\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_p\le <!--\; \le \; -->1\}$ be the ${\mathrm{\ell <!--\; \ell \; -->}}_p$ unit-ball in ${\mathbb{R}}^n$. Fix $h\in <!--\; \in \; -->[0,1]$.
Of all convex subsets A of ${\mathbb{B}}_{n,p}$ with $d(0,A)=h$, which shape maximizes the volume ?

What's the probability that Abe will win the dice game?
Abe and Bill are playing a game. A die is rolled each turn.
If the die lands 1 or 2, then Abe wins.
If the die lands 3, 4, or 5, then Bill wins.
If the die lands 6, another turn occurs.
What's the probability that Abe will win the game?
I think that the probability is $\frac{2}{5}$ just by counting the number of ways for Abe to win. I'm not sure how to formalize this though in terms of a geometric distribution.

n points are chosen randomly from a circumference of a circle. Find the probability that all points are on the same half of the circle. My intuition was that the probability is $1/2^{n-<!--\; -\; -->1}$ since if the first point is placed somewhere on the circumference, for each point there is probability of 1/2 to be placed on the half to the right of it and 1/2 to the left..but i feel i might be counting some probabilitys twice.

Probability of $X<Y$ for two independent geometric random variables, intuitively
Given two independent, geometric random variables X and Y with the same success probability $p\in <!--\; \in \; -->(0,1)$, what is $\mathbf{P}(X<Y)$?
The answer can be computed formally using total probability and one gets $\mathbf{P}(X<Y)=\frac{1-<!--\; -\; -->p}{2-<!--\; -\; -->p}$

Probability given by geometric model: $P(X=k)=p(1-<!--\; -\; -->p{)}^k$, calculate $P(X>1|X\le <!--\; \le \; -->2)$
Being $X\sim <!--\; \sim \; -->G(0.4)$, calculate:
a) $P(X=3)$
c) $P(X>1|X\le <!--\; \le \; -->2)$

Probability of ending with 2 blue balls in a row after n draws
The scenario is, we're playing a game where we have in a box 9 balls: 4 red, 3 green, 2 blue.
- We draw one ball at a time, and note down its color and then return it for the next draw, the game stops only when we have drawn the same color twice in a row.
- What is the probability of the game ending after n draws? $P(X=n)$?
- What is the probability of the last two balls being blue?

Geometric interpretation of multiplication of probabilities?
When dealing with abstract probability space $\mathrm{\Omega <!--\; \Omega \; -->}$ which consists of atomic events with measure ($P:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->\mathbb{R}$) defined for them, it seems natural to start immediately imagining simple cases like this: $\mathrm{\Omega <!--\; \Omega \; -->}$ is some closed area in 2-D space partitioned into subareas, representing atomic events. Measure is the area. The total area of $\mathrm{\Omega <!--\; \Omega \; -->}$ is 1.
In fact this is what they sometimes picture in textbooks with the help of vienne diagrams or whatever.
This intuition works fine for simple cases, especially with adding probabilities.
But then it comes to probability multiplication: I don't understand how to interpret it within this simple model. Is there a way?.. Is there at all some mental model to think about probability multiplication in simple geometric terms, besides Lebesgue terms?

CDF and Survival Function of Geometric Distribution
I am having trouble understanding the intuition behind the CDF and survival probability of a geometric distribution on both {0,1,…} and on {1,2,3,…}.
I know that a geometric starting from 0 is the number of failures before the first success and the geometric starting at 1 is the number of failures including the first success.
Can someone please explain the intuition behind the CDF and survival functions and explain what the formula is?
I think that the part confusing me is: Why is the following not true?
$p(X>n)={(1-<!--\; -\; -->p)}^{n+1}+{(1-<!--\; -\; -->p)}^{n+2}+...$ for a geometric starting at 0.

The Probability the number of flips of a fair coin to achieve 3 heads is mod 3
Compute the probability that the number of flips of a fair coin needed in order to achieve a total of exactly 3 Heads will be a multiple of 3.

Density of the first k coordinates of a uniform random variable
Suppose that X is distributed uniformly in the n-sphere $\sqrt{n}{\mathbf{S}}^{n-<!--\; -\; -->1}\subset <!--\; \subset \; -->{\mathbf{R}}^n$. Then apparently the distribution of $(X_1,\dots <!--\; \dots \; -->,X_k)$, the first $k<n$ coordinates of X has density $p(x_1,\dots <!--\; \dots \; -->,x_k)$ with respect to Lebesgue measure in ${\mathbf{R}}^k$, moreover if $r^2=x_1^2+\cdots <!--\; \cdots \; -->+x_k^2$, then it is proportional to
${(1-<!--\; -\; -->\frac{r^2}{n})}^{(n-<!--\; -\; -->k)/2-<!--\; -\; -->1},\text{if}0\le <!--\; \le \; -->r^2\le <!--\; \le \; -->n,$
and otherwise is 0. I tried to compute this using the fact that $(X_1,\dots <!--\; \dots \; -->,X_k)\stackrel{\mathrm{d}}{=}\sqrt{n}(g_1,\dots <!--\; \dots \; -->,g_k)/\sqrt{g_1^2+\cdots <!--\; \cdots \; -->+g_n^2}$, when $g_i$ are iid standard normal variables, but was unable to simplify the integrals. Does anyone know/can point me to a place where this density is derived?

Suppose a real number is chosen from the interval [-1, 1].
Find the probability that the number is positive, and using geometric probability, I get 1/2. Simple, right?
However, the next question asks what the probability of choosing a nonnegative real number is, and I do understand it is also 1/2
After finishing my homework, I'm wondering: Is it possible to choose 0, and what would the probability of choosing 0 be?

If X is a nonnegative $\mathrm{\sigma <!--\; \sigma \; -->}$-subGaussian random variable with $P(X=0)\ge <!--\; \ge \; -->p$, what is a good upper bound for $P(X\ge <!--\; \ge \; -->h)$?

Given an item with a value of X, it has a 50% chance to split/double, so now we have two items, each with value X. They each have separate, and half the chance to split as their parent object, so each one has in this case 25% chance to split. How can I calculate the average number of items, given that the initial chance to split is instead Y, and assuming it can theoretically split forever, and if the splits is also capped to some number N. Maybe format from 1 to N, where N could be infinity?

How to interpret the mean of geometric distribution
When I read my textbook, I find one conclusion which states that if the mean of geometric random variable is finite, then with probability 1, we can get the first success within "finite" step.
I am confused about the "finite" part. Why there is no chance that we will never get success? Since the mean of geometric random variable is 1/p, doesn't it mean we can always get success within finite experiments if $p>0$?

Let $R_1,...,R_n$ be independent continuous uniform over [0, 1] random variables.
The geometric mean of $R_1,...,R_n$ is defined by
$G_n=􏰠\sqrt[n]{R_1\times ...\times R_n}=(R_1\times ...\times R_n{)}^{\frac{1}{n}}.$
Show that $G_n$ converges in probability as well as with probability 1 (i.e. almost surely) to a constant, and identify the limit. Make sure you state any theorem you use.

Conditional probability with geometric distribution
X and Y are linearly independent random variables with geometric distribution. They have the same p parameter. How can I calculate the following expressions:
a) $P(X=k|X<Y)=?$
b) $P(X=k|X=Y)=?$
c) $\mathbb{E}(X\mid X+Y)=?$

Geometric probability (dealing with integer numbers)
Two numbers a and b are chosen from {1,2,…N} with replacement. Let $p_N$ be the probability that $a^2+b^2\le <!--\; \le \; -->N^2$. What is the value of $\underset{N\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{p}_N$?
With real numbers, I would just use the geometric approach: $\frac{\mathrm{\pi <!--\; \pi \; -->}\cdot <!--\; \cdot \; -->N^2/4}{N^2}$ so the limit as $N\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ equals $\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$.

In a vase a 10 balls: 4 red and 6 white. They are taken out one by one without replacement.
Let X be the stochastic variable that denotes how often we need to draw balls before we draw a white ball.
Calculate the probability function of X.

Probability in higher dimension
There are n non negative numbers $x_i<1$. It is given that $S_1=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i=1$. What is the probability that
$S_2=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i^2<c$ where
$\frac{1}{n}<c<1$
Assume that each $x_i$ is uniformly distributed.
Now I tried looking at this problem in terms of integrals:
$P(squares|sum)=\frac{P(squares\cap <!--\; \cap \; -->sum)}{P(sum)}=\frac{something}{{\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x_n}{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x_3-<!--\; -\; -->x_4-<!--\; -\; -->...-<!--\; -\; -->x_n}...{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x_2-<!--\; -\; -->x_3-<!--\; -\; -->...-<!--\; -\; -->x_n}d{x}_{1}d{x}_2...d{x}_{n-<!--\; -\; -->1}d{x}_n}$
Now for the numerator, if we substitute the constraint in the inequality, we get
$S_2+(1-<!--\; -\; -->S_1{)}^2=c$
$S_2+(1-<!--\; -\; -->S_1{)}^2=c$
Now how to get the numerator integral, I'm not quite sure.
Also, is there any better, more elegant way of doing this problem? (I thought of using vectors, where the problem can be written as: $\stackrel{\to <!--\; \to \; -->}{x}.\stackrel{\to <!--\; \to \; -->}{1}=1$ then $P(||\stackrel{\to <!--\; \to \; -->}{x}||<\sqrt{c})=?$. But then we would have to model the distribution of $||\stackrel{\to <!--\; \to \; -->}{x}||$ which I have no idea about)
Or maybe a more direct approach? Like the ratio of areas (or volumes?).

Branching Process - Extinction probability geometric
Consider a branching process with offspring distribution Geometric($\mathrm{\alpha <!--\; \alpha \; -->}$); that is, $p_k=\alpha (1-\alpha {)}^k$ for $k\ge 0$.
a) For what values of $\alpha \in (0,1)$ is the extinction probability $q=1$.
Let $\{Z_n{\}}_{n\ge <!--\; \ge \; -->0}$ be a branching process with $p_0>0$. Let $\mathrm{\mu <!--\; \mu \; -->}=\underset{k\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}k{p}_k$ be the mean of the offspring distribution and let $g(s)=\underset{k\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}{s}^k{p}_k$ be the probability generating function of the offspring distribution.
- If $\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->1$, then the extinction probability $q=1$.
- If $\mathrm{\mu <!--\; \mu \; -->}>1$, then the extinction probability q is the unique solution to the equation $s=g(s)$ with $s\in <!--\; \in \; -->(0,1)$.
b) Use the following proposition to give a formula for the extinction probability of the branching process for any value of the parameter $\alpha \in (0,1)$.